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Let \(\mathbf{m}\) be a vector of magnitude \(\sqrt{3}\) and perpendicular to the vectors \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\) and \(\hat{\mathbf{j}}-\hat{\mathbf{k}}\). Let \(\mathbf{n}\) be another vector of magnitude \(2 \sqrt{6}\) and perpendicular to the vectors \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}\) and \(\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\). The area (in sq. units) of the triangle formed with \(\mathbf{m}\) and \(\mathbf{n}\) as sides is
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1845 Upvotes
Verified Answer
The correct answer is:
\(3 \sqrt{2}\)
Given,
\(\mathbf{m}=\sqrt{3} \times \text { unit vectors of }[(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}-\hat{\mathbf{k}})]\)
\(\begin{aligned}
\because(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}-\hat{\mathbf{k}}) & =\left[\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & 1 & 0 \\
0 & 1 & -1
\end{array}\right] \\
& =\hat{\mathbf{i}}(-1-0)-\hat{\mathbf{j}}(-1-0)+\hat{\mathbf{k}}(\mathbf{l}) \\
& =-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\
\therefore \quad \mathbf{m} & =\sqrt{3} \times \frac{-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{1+1+1}} \\
& =\sqrt{3} \times \frac{-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}
\end{aligned}\)
and \(\mathbf{n}=2 \sqrt{6} \times\) unit vectors of \([(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}+2 \hat{\mathbf{k}})]\)
\(\begin{aligned}
& \therefore(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\
&=\left[\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
2 & -1 & 0 \\
0 & 1 & 2
\end{array}\right] \\
&=\hat{\mathbf{i}}(-2-0)-\hat{\mathbf{j}}(4-0)+\hat{\mathbf{k}}(2+0)=-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
& \mathbf{n}=2 \sqrt{6} \times \frac{-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{\sqrt{4+16+4}}=2 \sqrt{6} \times \frac{-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{2 \sqrt{6}}
\end{aligned}\)
\(=-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\)
Now, required area
\(\begin{aligned}
& =\frac{1}{2}(\mathbf{m} \times \mathbf{n})=\frac{1}{2}|(-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \times(-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})| \\
& =\frac{1}{2}\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
-1 & 1 & 1 \\
-2 & -4 & 2
\end{array}\right| \\
& =\frac{1}{2} \times \mid \hat{\mathbf{i}}(2+4)-\hat{\mathbf{j}}(-2+2)+\hat{\mathbf{k}}(4+2|| \\
& =\frac{1}{2} \times|6 \hat{\mathbf{i}}+6 \hat{\mathbf{k}}|=\frac{1}{2} \sqrt{6^2+6^2}=\frac{1}{2} \times 6 \times \sqrt{2}=3 \sqrt{2}
\end{aligned}\)
\(\mathbf{m}=\sqrt{3} \times \text { unit vectors of }[(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}-\hat{\mathbf{k}})]\)
\(\begin{aligned}
\because(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}-\hat{\mathbf{k}}) & =\left[\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & 1 & 0 \\
0 & 1 & -1
\end{array}\right] \\
& =\hat{\mathbf{i}}(-1-0)-\hat{\mathbf{j}}(-1-0)+\hat{\mathbf{k}}(\mathbf{l}) \\
& =-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\
\therefore \quad \mathbf{m} & =\sqrt{3} \times \frac{-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{1+1+1}} \\
& =\sqrt{3} \times \frac{-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}
\end{aligned}\)
and \(\mathbf{n}=2 \sqrt{6} \times\) unit vectors of \([(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}+2 \hat{\mathbf{k}})]\)
\(\begin{aligned}
& \therefore(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}) \times(\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\
&=\left[\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
2 & -1 & 0 \\
0 & 1 & 2
\end{array}\right] \\
&=\hat{\mathbf{i}}(-2-0)-\hat{\mathbf{j}}(4-0)+\hat{\mathbf{k}}(2+0)=-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
& \mathbf{n}=2 \sqrt{6} \times \frac{-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{\sqrt{4+16+4}}=2 \sqrt{6} \times \frac{-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{2 \sqrt{6}}
\end{aligned}\)
\(=-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\)
Now, required area
\(\begin{aligned}
& =\frac{1}{2}(\mathbf{m} \times \mathbf{n})=\frac{1}{2}|(-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \times(-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})| \\
& =\frac{1}{2}\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
-1 & 1 & 1 \\
-2 & -4 & 2
\end{array}\right| \\
& =\frac{1}{2} \times \mid \hat{\mathbf{i}}(2+4)-\hat{\mathbf{j}}(-2+2)+\hat{\mathbf{k}}(4+2|| \\
& =\frac{1}{2} \times|6 \hat{\mathbf{i}}+6 \hat{\mathbf{k}}|=\frac{1}{2} \sqrt{6^2+6^2}=\frac{1}{2} \times 6 \times \sqrt{2}=3 \sqrt{2}
\end{aligned}\)
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